If `f'(x)=x-3/x^2,f(1)=11/2`, find f(x).

Answer

To find the function \( f(x) \) given that \( f'(x) = \frac{x - 3}{x^2} \) and \( f(1) = \frac{11}{2} \), we will follow these steps:

### Step 1: Integrate \( f'(x) \)

We start with the derivative:

\[ f'(x) = \frac{x - 3}{x^2} \]

We can rewrite this as:

\[ f'(x) = \frac{x}{x^2} - \frac{3}{x^2} = \frac{1}{x} - \frac{3}{x^2} \]

Now, we integrate \( f'(x) \):

\[ f(x) = \int \left( \frac{1}{x} - \frac{3}{x^2} \right) dx \]

### Step 2: Perform the Integration

Integrating term by term:

1. The integral of \( \frac{1}{x} \) is \( \ln |x| \).
2. The integral of \( -\frac{3}{x^2} \) is \( 3 \cdot \frac{1}{x} \) (since \( \int x^{-2} dx = -\frac{1}{x} \)).

Thus, we have:

\[ f(x) = \ln |x| + 3 \cdot \left(-\frac{1}{x}\right) + C = \ln |x| + \frac{3}{x} + C \]

### Step 3: Use the Initial Condition to Find \( C \)

We know that \( f(1) = \frac{11}{2} \). Plugging in \( x = 1 \):

\[ f(1) = \ln |1| + \frac{3}{1} + C = 0 + 3 + C = C + 3 \]

Setting this equal to \( \frac{11}{2} \):

\[ C + 3 = \frac{11}{2} \]

### Step 4: Solve for \( C \)

To find \( C \):

\[ C = \frac{11}{2} - 3 = \frac{11}{2} - \frac{6}{2} = \frac{5}{2} \]

### Step 5: Write the Final Expression for \( f(x) \)

Now substituting \( C \) back into our expression for \( f(x) \):

\[ f(x) = \ln |x| + \frac{3}{x} + \frac{5}{2} \]

Thus, the final answer is:

\[ \boxed{f(x) = \ln |x| + \frac{3}{x} + \frac{5}{2}} \]